Find the minimum value of the function f(x) = -x^3 + 3ax on the interval 0 ≤ x ≤ 2. assume that a > 0. give the minimum value in terms of a.

for 0 ≤ a < , minimum is -8 + 6a

              for ≤ a ≤ 2, minimum is 0

Step-by-step explanation:

f(x) = -x³ + 3ax is continuous and differentiable on the interval [0, 2], so the minimum will occur at a critical point or an endpoint.

Step 1: Find the critical point(s) between [0, 2] by setting the derivative of the function equal to zero and solving for x.

f(x) = -x³ + 3ax

f'(x) = -3x² + 3a

0 = -3x² + 3a

3x² = 3a

x² = a

x = ±√a

x = √a   and   x = -√a

                          ↓

                      not valid because it is not in the interval [0, 2]

So, the only critical point is at x = √a

Step 2: Evaluate f(x) at the critical point (x = √a) and endpoints (x = 0 and x = 2)

f(x) = -x³ + 3ax

f(0) = -(0)³ + 3a(0)

     =  0 + 0

     = 0

f(√a) = -(√a)³ + 3a(√a)

        = -a√a + 3a√a

        = 2a√a

f(2) = -(2)³ + 3a(2)

     =  -8 + 6a

Step 3: Evaluate the values of "a" that will determine the minimum

Notice that f(a) > f(2) for all a-values so this can be disregarded for the minimum. So, we need to see when f(2) < f(0)

-8 + 6a < 0

       6a < 8

Thus, the minimum is -8 + 6a when "a" is between 0 (the endpoint) and .  The minimum is 0 when "a" is between and 2 (the endpoint).

Bonus: See the attached graph to see minimum for different values of "a".

as "a" approaches 0, y = -8 is the lowest value

when a = 1,  y = -2 is the lowest value

when a = ,  y = 0 is the lowest value

when a = 2,  y = 0 is the lowest value

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

i think the correct option is the second option.